TWO VERTICES OF AN ISOSCELES TRIANGLE IS (2,0) AND (2,5).FIND IT’S THIRD VERTEX.. NO SPAMMING SPAMMING MAY LEAD TO DELETION OF UR ACCOUNT..

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TWO VERTICES OF AN ISOSCELES TRIANGLE IS (2,0) AND (2,5).FIND IT’S THIRD VERTEX.. NO SPAMMING SPAMMING MAY LEAD TO DELETION OF UR ACCOUNT.. THANK YOU!!

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Josie 1 month 2021-10-27T02:44:49+00:00 2 Answers 0 views 0

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    0
    2021-10-27T02:46:28+00:00

    Step-by-step explanation:

    hola sis here is ur ans

    Let ΔABC be the isosceles triangle, the third vertex be C(a,b).

    and A(2,0) and B(2,5)

    Let AC and BC be equal sides of the triangle

    By distance formula we have,answr

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    11th

    Maths

    Straight Lines

    Introduction

    Two vertices of an isoscele…

    MATHS

    Two vertices of an isosceles triangle are (2,0) and (2,5). Find the third vertex if the length of the equal sides is 3.

    MEDIUM

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    ANSWER

    Let ΔABC be the isosceles triangle, the third vertex be C(a,b).

    and A(2,0) and B(2,5)

    Let AC and BC be equal sides of the triangle

    By distance formula we have,

    D=

    (x

    2

    −x

    1

    )+(y

    2

    −y

    1

    )

    ∴AB=

    (2−2)

    2

    +(5−0)

    2

    =

    25

    =5

    Now,

    AC=3 [length of equal sides=3 ]

    AC

    2

    =3

    2

    =9……….(a)

    but by distance formula,

    AC

    2

    =

    (a−2)

    2

    +b

    2

    ………..(b)

    Combining (a) and (b)

    ⇒(a−2)

    2

    +b

    2

    =9

    ⇒a

    2

    +4−4a+b

    2

    =9

    ⇒a

    2

    +b

    2

    −4a=5 ———- (1)

    Consider BC

    BC

    2

    =

    (a−2)

    2

    +(b−5)

    2

    but BC=3

    ∴(a−2)

    2

    +(b−5)

    2

    =9

    ⇒a

    2

    +4−4a+b

    2

    +25−10b=9

    ⇒a

    2

    +b

    2

    −4a−10b+20=0 ——- (2)

    ⇒−10b+20+5=0 ———(from 1 )

    ⇒10b=25

    ⇒b=

    10

    25

    =

    2

    5

    Now, a

    2

    +

    4

    25

    −4a=5

    ⇒a

    2

    −4a=

    4

    −5

    ⇒a

    2

    −4a+

    4

    5

    =0

    ⇒a=

    2

    16−4∗1∗

    4

    5

    =

    2

    11

    =

    2

    2+

    11

    ,

    2

    2−

    11

    ∴ Coordinate of third vertex (a,b) = (

    2

    2+

    11

    ,

    2

    5

    ) and (

    2

    2−

    11

    ,

    2

    5

    )

    0
    2021-10-27T02:46:36+00:00

    Answer

    if it a equilateral triangle

    given vertex A = (2,0) B=(2,5)

    AB=AC=BC=5cm

    as AC=5cm

    vertex C will be (2,0)+(5,0)=(7,0)

    so the third vertex is (7,0)

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